# Algebraic manipulation of binomial theorem

Prove, by algebraic manipulation, that:

${{2n} \choose {n}} + {{2n} \choose {n+1}}={1\over2} {{2n+2} \choose {n+1}}$

-
I don't really like the notation used, and its not typed using math tex, if you could refine your question perhaps with binomial coeifficents, typed correctly, I could help you – Ethan Dec 9 '12 at 3:31
This just looks like a special case of pascals identity, Jillian you should slow down on the question posting and try to research some of these yourself, you could have found an algebraic proof on wikipedia, which I found after several searchs – Ethan Dec 9 '12 at 3:35
If you're going to go on a Math forum just to bring people down, perhaps rethink why you're here. – Jillian O'Reilly Dec 9 '12 at 15:32
There's a button called "Edit" @Ethan – xiamx Dec 11 '12 at 0:50

$$\binom{2n}{n}+\binom{2n}{n+1}=\frac{(2n)!}{(n!)^2}+\frac{(2n)!}{(n+1)!(n-1)!}=\frac{(2n)!}{((n-1)!)^2}\left(\frac{1}{n^2}+\frac{1}{n(n+1)}\right)=$$

$$=\frac{(2n)!}{((n-1)!)^2}\frac{2n+1}{n^2(n+1)}=\frac{(2n+1)!}{n!(n+1)!}$$

Try to complete the few steps left now...

-

A general method for such identities, similar to what DonAntonio wrote but requiring less thought, is to write the binomial coefficients as fractions of factorials and clear all the denominators. (In the case at hand you'll multiply by $2((n+1)!)^2$.) Then cancel the biggest common factors you can find. What's left should be fairly easy to compute.

For this particular problem, though, a nicer-looking (but perhaps harder to find) alternative is essentially what Ethan suggested in a comment. Multiply both sides of the proposed equation by 2. (If you get the impression that I really don't like fractions, you're right.) On the left, use the fact that $C(2n,n+1)=C(2n,n-1)$ to replace one of the two $C(2n,n+1)$'s. That will set you up for three applications of Pascal's identity, which convert the left side to the right.

-

$$\binom m r+\binom m{r+1}=\frac{m!}{(m-r)! r!}+\frac{m!}{(m-r-1)! (r+1)!}=\frac{r+1}{m+1}\frac{(m+1)!}{\{(m+1)-(r+1)\}! (r+1)!}+\frac{m-r}{m+1}\frac{(m+1)!}{\{m-(r+1)\}! (r+1)!}$$ $$=\binom{m+1}{r+1}\cdot\left(\frac{r+1}{m+1}+\frac{m-r}{m+1}\right)=\binom{m+1}{r+1}$$

Putting $m=2n,r=n,$

$$\binom {2n} n+\binom {2n}{n+1}=\binom{2n+1}{n+1}=\frac{(2n+1)!}{n!(n+1)!}=\frac1 2 \frac{(2n+2)\cdot (2n+1)!}{(n+1)\cdot n!(n+1)!}=\frac1 2\frac{(2n+2)!}{(n+1)!(n+1)!}=\frac1 2\binom{2n+2}{n+1}$$

-